Undergrad Discrepancy in Co-Moving Charges Magnetic Force?

[Dale]

Why does this factor not lead to a different B field? If the proof is complex maybe you can just post a link, that's ok.

$\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}}$

The easiest way to see that is to just look at the retarded potentials. In the retarded potentials A depends only on J and B depends only on A. So if J is identical then B is also identical.

 

[greswd]

The easiest way to see that is to just look at the retarded potentials. In the retarded potentials A depends only on J and B depends only on A. So if J is identical then B is also identical.

Wow that's so counter-intuitive. I think almost anyone looking at the problem at first glance would assume that the B-field would be affected.

Does this mean that the relativistic changes within the charge distribution somehow offset each other?

 

[Dale]

Does this mean that the relativistic changes within the charge distribution somehow offset each other?

I don't know what you mean by that. Aren't you dealing with a single frame? So what relativistic changes are you talking about in a single frame?

 

[greswd]

what relativistic changes are you talking about in a single frame?

The $\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}}$ factor getting larger and larger as the speeds of the charge distribution increase.

The increase in speed is matched by a decrease in charge density, as mentioned in #28, so that J remains unchanged.

 

[Dale]

The $\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}}$ factor getting larger and larger as the speeds of the charge distribution increase.

I am not exactly sure what that factor is or where it comes from (in terms of assumptions in the derivation). If it is for a point charge then you cannot increase v without increasing J also.

 

[greswd]

I am not exactly sure what that factor is or where it comes from (in terms of assumptions in the derivation).

I first mentioned it in #16:

here is the formula I wanted to post:

$\mathbf{B} =\frac{\mu_0 q}{4\pi}\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}} \mathbf{v} \times \frac{\mathbf{\hat r'}}{|\mathbf r'|^2}$
from https://en.wikipedia.org/wiki/Biot–Savart_law#Point_charge_at_constant_velocity

Is it considered relativistically accurate?

I thought you had acknowledged it in #31 as well:

Why does this factor not lead to a different B field? If the proof is complex maybe you can just post a link, that's ok.

$\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}}$

The easiest way to see that is to just look at the retarded potentials. In the retarded potentials A depends only on J and B depends only on A. So if J is identical then B is also identical.

If it is for a point charge then you cannot increase v without increasing J also.

A continuous charge distribution can be approximated by a 'cloud' or 'fine mist' of point charges.

I wrote:

Imagine two fields of charge. One has a high charge density but slow speeds. One has very low charge density but high speeds, relativistic speeds.

Therefore both have the same J fields, numerically speaking.

I am talking about the comparison of two different scenarios, one with dense charge and slow speeds, and another with less dense charge and faster speeds.

 

[Dale]

I thought you had acknowledged it in #31 as well

I was just quoting your post.

A continuous charge distribution can be approximated by a 'cloud' or 'fine mist' of point charges

You can get a lot of incorrect intuition that way. For example, you might conclude that you should get synchrotron radiation from a DC current loop.

I am talking about two different scenarios, one with dense charge and slow speeds, and another with less dense charge and faster speeds.

If J is the same in the two different scenarios then B is the same in the two different scenarios.

 

[greswd]

You can get a lot of incorrect intuition that way. For example, you might conclude that you should get synchrotron radiation from a DC current loop.

haha, that's interesting, why can't we get sync radiation from a current loop? Isn't the current in a loop also made of moving charged particles?

If J is the same in the two different scenarios then B is the same in the two different scenarios.

Yes, you told me that already. So I asked if this was due to the relativistic effects offsetting each other.

The easiest way to see that is to just look at the retarded potentials. In the retarded potentials A depends only on J and B depends only on A. So if J is identical then B is also identical.

Wow that's so counter-intuitive. I think almost anyone looking at the problem at first glance would assume that the B-field would be affected.

Does this mean that the relativistic changes within the charge distribution somehow offset each other?

I asked because B being the same in both scenarios is so counter-intuitive to me.

 

[Dale]

why can't we get sync radiation from a current loop?

Because J is not a function of time for a DC current loop so there is no radiation.

Isn't the current in a loop also made of moving charged particles?

Which is precisely why thinking of a continuous current distribution as a collection of charged point particles can give you faulty intuition.

Yes, you told me that already. So I asked if this was due to the relativistic effects offsetting each other

Understood. Unfortunately, what you mean by "relativistic effects" here is a term from an equation for point charges, so it doesn't readily apply for continuous charge distributions.

Of course, it is possible to go from one to the other through careful integration, but it is not trivial and the proof from continuous distributions is so easy that I don't see the point in going through that exercise.

 

[greswd]

Because J is not a function of time for a DC current loop so there is no radiation.

Which is precisely why thinking of a continuous current distribution as a collection of charged point particles can give you faulty intuition.

Understood. Unfortunately, what you mean by "relativistic effects" here is a term from an equation for point charges, so it doesn't readily apply for continuous charge distributions.

Of course, it is possible to go from one to the other through careful integration, but it is not trivial and the proof from continuous distributions is so easy that I don't see the point in going through that exercise.

Do you know of any webpage that shows how to go from one to the other?I found this article about synchrotron radiation: http://puhep1.princeton.edu/~kirkmcd/examples/steadycurrent.pdf

If the steady current were a continuous DC current, all of its multipole moments would be constant, and we would expect no radiation.1 For a current consisting of electrons, it must be that the radiation is canceled by destructive interference between the radiation fields of the large number N of electrons that make up the steady current.

I sound like a broken record; is it useful to think of the relativistic effects offsetting one another just like the destructive interference between the radiation fields of a continuous steady current?

 

[Dale]

Do you know of any webpage that shows how to go from one to the other?

Not for this specific problem, no. But you can get a general idea of the approach in calculus based discussions of Newton's shell theorem.

I found this article about synchrotron radiation: http://puhep1.princeton.edu/~kirkmcd/examples/steadycurrent.pdf

The comment about destructive interference is perfect.

is it useful to think of the relativistic effects offsetting one another

I can't answer this for the reason described above.

 

[greswd]

I can't answer this for the reason described above.

which reason?

 

[Dale]

which reason?

Last two paragraphs of post 39

 

[greswd]

Last two paragraphs of post 39

so its because the proof is not easy? ok.

 

[Dale]

so its because the proof is not easy? ok.

Well, the proof which I gave is very easy, it just doesn't involve the terms you are choosing to focus on. Putting those terms in would be hard and would serve no other purpose, so I am not inclined to do it nor do I know of anyone else who was. You are certainly welcome to do it if it feels important to you.

 

[greswd]

Well, the proof which I gave is very easy, it just doesn't involve the terms you are choosing to focus on. Putting those terms in would be hard and would serve no other purpose, so I am not inclined to do it nor do I know of anyone else who was. You are certainly welcome to do it if it feels important to you.

Yeah, no worries. You've already explained how it works conceptually, which has been enough to answer my question(s).

It is actually important to me, but since its so hard I'll save it for another day.

Thanks for all the help rendered for this thread.